0

这是一个UDF函数要求：

CREATE FUNCTION [dbo].[ConditionEvaluation]

(@condEquation VARCHAR(MAX),
@condParameters VARCHAR(MAX
)
RETURNS bit
AS

@condEquation和的值@condParameters是

  • 示例 1：@condEquation = (C01 & C02)和@condParameters =10
  • 示例 2：@condEquation = ((C01 & C02) | C3)和@condParameters =101

因此对于每个 C 系列条件，@condEquation第二个参数中将提供相应的 0 或 1 值，即@condParameters。

我想将上述条件评估为...

  • 示例 1：select (1 & 0)
  • 示例 2：select ((1 & 0) | 0)

参数@condEquation可能包含方程中任意数量的 C。但是参数2中会有相应的位数。

我这里利用了 SQL Select 语句的条件求值能力，想将求值结果返回为 0 或 1。

如何使用 UDF 做到这一点？

4

2 回答 2

0

为了做到这一点，我不得不做两个功能：

create function [dbo].[f_test1](@CondEquation varchar(50))
returns bit
as
begin
declare @result bit
;with a as
(
select replace(replace(replace(@CondEquation, '(',''), ')',''), ' ','') n
),
b as
(
select n, 1 rn from a
union all
select stuff(n, patindex('%&%', n) - 1, 3 , case when substring(n, patindex('%&%', n) - 1, 3) like '%0%' then 0 else 1 end), rn+1
from b
where patindex('%&%', n)> 0
), c as
(
select n from (
select n, row_number() over (order by rn desc) rn2 from b
) a where rn2 = 1
), d as
(
select n, 1 rn from c
union all
select stuff(n, patindex('%|%', n) - 1, 3 , case when substring(n, patindex('%|%', n) - 1, 3) like '%1%' then 1 else 0 end), rn+1
from d
where patindex('%|%', n)> 0
), e as
(
select n from (
select n, row_number() over (order by rn desc) rn2 from d
) a where rn2 = 1
)
select @result=n from e
return @result
end
go

create function [dbo].[f_test2](@CondEquation varchar(max), @condparameters varchar(max))
returns bit
as
begin
declare @result bit

declare @i int = 1
while @i <= len(@condparameters)
begin
set @CondEquation = replace(@CondEquation, 'c' + right(@i+100, 2), substring(@condparameters, @i, 1))
set @i += 1
end

declare @returnvalue bit
;with a as
(
select @CondEquation a, 1 rn
union all
select stuff(a.a, z.a-z.b+1, z.b+1,[dbo].[f_test1](substring(a.a, z.a-z.b+1, z.b+1)) ), rn+1  from
a cross apply(
select patindex('%_)%', a.a) a, charindex('(', reverse(left(a.a, patindex('%_)%', a.a)))) b
where a.a like '%)%'
) z
), b as
(
select a, row_number() over (order by rn desc) rn from a
)
select @returnvalue = [dbo].[f_test1](a) from b where rn = 1

return @returnvalue
end
go

您可以像这样测试功能

select dbo.[f_test2]('((C01 & C02) | C03)', '110')

请注意，使用正确的参数格式很重要。你不能写 C3 而不是 C03。我强烈建议您回去评估您的旧问题以获得正确答案。

于 2012-07-31T15:42:00.687 回答
0 
    CREATE FUNCTION [dbo].[ConditionEvaluation] 

(@condEquation VARCHAR(MAX), 
@condParameters VARCHAR(MAX) 
) 
RETURNS bit 
AS 
BEGIN

declare @len int=len(@condParameters)
declare @val varchar(100)
declare @i int=1
declare @Sindex int=0
declare @op bit
set @condEquation=replace(replace(@condEquation,' ',','),')',',)')

while(@i<=@len )
begin
set @val=SUBSTRING (@condParameters,@i,1)
set @Sindex =CHARINDEX ('C',@condEquation)
set @condEquation=stuff(@condEquation,@Sindex ,charindex(',',@condEquation)-2 ,@val)
set @i=@i+1
end

set @condEquation= 'select @OP1'+replace(@condEquation,',','')+' as output1'


execute sp_executesql @condEquation,N'@OP1 int OUTPUT',@OP1=@OP OUTPUT

return @OP

END
于 2012-07-31T17:38:33.933 回答